Week #8 - Optimization and Newton s Method Section 4.5

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1 Week #8 - Optimization and Newton s Method Section 4.5 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS 1. The bending moment M of a beam, supported at one end, at a distance from the support is given by M = 1 2 wl 1 2 w2 where L is the length of the beam, and w is the uniform load per unit length. Find the point on the beam where the moment is greatest. We look for critical points of M: dm d = 1 wl w 2 Now dm/d = 0 when = L/2. At this point d 2 M/d 2 = w so this point is a local maimum. The graph of M() is a parabola opening downward, so the local maimum is also the global maimum. 7. The efficiency of a screw, E, is given by E = θ µθ2 µ + θ where θ is the angle of pitch of the thread and µ is the coefficient of friction of the material, a (positive) constant. What value of θ maimizes E? Differentiating, to look for critical points: de dθ = µ + θ(1 2µθ) (θ µθ2 ) (µ + θ) 2 = µ(1 2µθ θ2 ) (µ + θ) 2 de/dθ = 0 when θ = µ ± 1 + µ 2. Since θ > 0, the only possible critical point is when θ = µ µ 2. Slopes go from positive (at θ = 0 to negative after the critical point, so it is a local maimum. Because of the positive slopes to the left and negative slopes to the right of the critical the local maimum must be the global maimum. 1

2 11. Figure 4.69 shows the curves y =, = 9, y = 0, and a rectangle with its sides parallel to the aes and its left end at = a. Find the dimensions of the rectangle having the maimum possible area. Figure 4.69 We wish to choose a to maimize the area of the rectangle with corners at (a, a) and (9, a). The area of this rectangle will be given by the formula R = h l = a(9 a) = 9a 1/2 a 3/2 We are restricted to 0 a 9. To maimize this area, we set dr/da = 0, and then check that the resulting area is greater than the area if a = 0 or a = 9. Since R = 0 if a = 0 or a = 9, all we need to do is to find where dr/da = 0: dr da = 9 2 a 1/2 3 2 a1/2 = a = 3 a 2 18 = 6a a = 3 Thus the dimensions of the maimal rectangle are 6 by If you have 100 feet of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose? Let w and l be the width and length, respectively, of the rectangular area you wish to enclose. Then length = w + w + l = 100 feet l = 100 2w Using that relationship, we can simplify the area to be a function of only w. Area = w l = w(100 2w) = 100w 2w 2 To maimize the area, we look for the critical point of the area. If we let the area be representing A(w), 2

3 A (w) = 100 4w Setting the derivative equal to zero, 4w = 100 or w = 25. Because the area function is a parabola, opening downwards (negative coefficient of w 2 in A(w)), we know that this represents the global maimum of the area. Thus the maimum area of A(25) = w (100 2w) = = 1250 square feet is achieved when w = 25 feet. 34. To get the best view of the Statue of Liberty in Figure 4.76, you should be at the position where θ is a maimum. If the statue stands 92 meters high, including the pedestal, which is 46 meters high, how far from the base should you be? Hint: [Find a formula for θ in terms of your distance from the base. Use this function to maimize θ, noting that 0 θ π/2.] Figure 4.76 We want to maimize the viewing angle, which is θ = θ 1 θ 2. Optimizing this requires some careful trig work. If the distance from the boat to the base 3

4 of the statue is, then tan(θ 1 ) = 92 tan(θ 2 ) = 46 ( ) 92 so θ 1 = arctan ( ) 46 so θ 1 = arctan Then θ = θ 1 θ 2 = arctan ( ) ( ) arctan for > 0 We look for critical points of the function by computing dθ/d, remembering that d d arctan() = : dθ d = (92/) 2 = ( ) ( ) (46/) 2 2 = 92( ) + 46( ) ( )( ) 46( ) = ( )( ) Setting this derivative equal to zero, 0 = 2 = ( ) ( )( ) = ± 4232 ±65.05 Since > 0, the critical point is = meters. To verify that this is indeed where θ attains a maimum, we not that dθ/d > 0 for 0 < < 4232 and dθ/d < 0 for > 4232, and so is a local minimum. Given the sign of the slopes on 0 < <, this must also be the global maimum. The best distance to stay view the statue is roughly 65 meters away from the base. QUIZ PREPARATION PROBLEMS 4

5. In a chemical reaction, substance A combines with substance B to form substance Y. At the start of the reaction, the quantity of A present is a grams, and the quantity of B present is b grams.

5 5. In a chemical reaction, substance A combines with substance B to form substance Y. At the start of the reaction, the quantity of A present is a grams, and the quantity of B present is b grams. Assume a < b. At time t seconds after the start of the reaction, the quantity of Y present is y grams. For certain types of reactions, the rate of the reaction, in grams/sec, is given by Rate = k(a y)(b y) (a) For what values of y is the rate nonnegative? Graph the rate against y. (b) Use your graph to find the value of y at which the rate of the reaction is fastest. (a) If we epect the rate to be nonnegative, we must have 0 y a and 0 y b. Since we assume a < b, we restrict y to 0 y a. In fact, the epression for the rate is nonnegative for y greater than b, but these values of y are not meaningful for the reaction. Rate 0 a b y (b) From the graph, we see that the maimum rate occurs when y = 0; that is, at the start of the reaction. 18. A rectangular beam is cut from a cylindrical log of radius 30 cm. The strength of a beam of width w and height h is proportional to wh 2. (See Figure 4.70.) Find the width and height of the beam of maimum strength. Figure 4.70 Since the beam diagonal length must always be 60, h and w must fit into the right angle triangle shown below: 5

6 60 h w Thus w 2 + h 2 = 60 2, or h 2 = 3600 w 2. We want to optimize the strength: S = kwh 2 replacing h 2 with 3600 w 2 : = kw(3600 w 2 ) = k(3600w w 3 ) Differentiating, setting S (w) = 0, S (w) = k(3600 3w 2 ) Setting S = 0: 0 = w 2 w 2 = 1200 w = ± 1200 ±34.64 Only the positive solution is feasible for this problem, so w = is the critical point of interest. To show this point is a global maimum for possible ranges of w, 0 w 60, we note that S(0) = 0 and S(60) = 0. Thus, a beam with dimensions w cm and l = 3600 w cm will have the optimal strength. 19. A landscape architect plans to enclose a 3000 square foot rectangular region in a botanical garden. She will use shrubs costing $25 per foot along three sides and fencing costing $10 per foot along the fourth side. Find the minimum total cost. y The total area is y = 3000, so y = 3000/. Suppose that the left and right edges and the lower edge have the shrubs, while the top edge has the fencing. Then the total cost function would be 6

7 Finding the critical points, C = 25(y + + y) + 10 = 50y + 35 C() = = 150, C () = 150, Setting C = 0: 0 = 150, = 150, ,000 30,000 = ± = ± 35 7 ±65.5 feet Only the positive root is feasible, as we are only interested in > 0, so is the only critical point of interest. To show this is a global minimum on the domain of the problem, we compare this with lim C() = + 0 lim C() = + Thus the critical point must be a global minimum, so 65.5 feet and y = 3000/ 48.5 feet give the dimensions of the yard with the lowest fencing/shrub cost, C(65.5) $ A light is suspended at a height h above the floor. (See Figure 4.71.) The illumination at the point P is inversely proportional to the square of the distance from the point P to the light and directly proportional to the cosine of the angle θ. How far from the floor should the light be to maimize the illumination at the point P? Figure 4.71 Since r 2 = y and cos θ = h/r = h/ h , we have 7

8 I = k cos θ = r 2 kh (h ) 3/2 To find the height at which I(h) is maimized, we look for critical points of I(h): di dh = k(h ) 3/2 ( 3 2 (h ) 1/2 2h)(kh) (h ) 3 Take out common factors: = k(h ) 1/2 [(h ) 3h 2 ] (h ) 3 Cancel (h ) 1/2 with denominator: = k (h ) 3h 2 (h ) 5/ h2 Simplify numerator: = k (h ) 5/2 Setting this equal to zero to identify critical points, and solving for h: 100 2h2 0 = k (h ) 5/2 100 = 2h 2 h = 50 Since di/dh > 0 for 0 h < 50, and di/dh < 0 for h > 50, we know that the light intensity is a maimum when h = A bird such as a starling feeds worms to its young. To collect worms, the bird flies to a site where worms are to be found, picks up several in its beak, and flies back to its nest. The loading curve in Figure 4.73 shows how the number of worms (the load) a starling collects depends on the time it has been searching for them.5 The curve is concave down because the bird can pick up worms more efficiently when its beak is empty; when its beak is partly full, the bird becomes much less efficient. The traveling time (from nest to site and back) is represented by the distance PO in Figure The bird wants to maimize the rate at which it brings worms to the nest, where Rate worms arrive = Load Traveling time + Searching time (a) Draw a line in Figure 4.73 whose slope is this rate. (b) Using the graph, estimate the load which maimizes this rate. (c) If the traveling time is increased, does the optimal load increase or decrease? Why? 8

9 Figure 4.73 (a) A possible line is shown on the graph below. Its slope is (load collected) / (total travel time), or the rate at which worms arrive back at the nest. (b) The slope of the line is maimized when the line is tangent to the loading curve, which happens with the black line on the graph below. (c) If the traveling time is increase, the point P moves to the left, the blue line on the graph below would show the new optimal slope. The contact point is higher, so the bird would be more heavily loaded, but would have a net lower efficiency, as the blue line has a lower slope than the earlier black line. 9

Week #8 - Optimization and Newton s Method Section 4.5

Week #8 - Optimization and Newton s Method Section 4.5 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 2005 by John Wiley & Sons, Inc. This material is used by permission